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Find the Maclaurin series for the function $\frac{1}{1-x^{2}}$ by using partial fractions or otherwise.
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The answer is obtained at once using the known series for $\frac{1}{1-x}$ and letting $x \rightarrow x^{2}$. With partial fractions, $\frac{1}{1-x^{2}}=\frac{1}{(1-x)(1+x)}=\frac{A}{1-x}+\frac{B}{1+x} .$ By the cover up method, $A=\frac{1}{2}, B=\frac{1}{2}$ so that
\begin{aligned} \frac{1}{1-x^{2}}=\frac{1}{2}\left(\frac{1}{1-x}+\frac{1}{1+x}\right) . \text { Since } & \\ & \frac{1}{1-x}=1+x+x^{2}+x^{3}+x^{4}+\cdots \\ & \frac{1}{1+x}=1-x+x^{2}-x^{3}+x^{4}+\cdots \end{aligned}
it follows that

\begin{aligned}
\frac{1}{1-x^{2}} &=\frac{1}{2}\left(2+2 x^{2}+2 x^{4}+\cdots\right) \\
&=1+x^{2}+x^{4}+\cdots
\end{aligned}
by Diamond (74,866 points)

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